/*
Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?


Note:
The solution is guaranteed to be unique.
*/

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <fstream>
#include <sstream>
#include <unordered_set>
#include "print.h"
using namespace std;

/**
* Definition for binary tree*/


void testForStack()
{
	stack<int> mystack;
	mystack.push(10);
	mystack.push(20);
	mystack.top() -= 5;
	cout << "mystack.top() is now " << mystack.top() << endl;
}

void testForIntToString()
{
	int a = 10;
	stringstream ss;
	ss << a;
	string str = ss.str();
	cout << str << endl;

	string str1 = to_string(a);

}



class Solution {
public:
	int singleNumber(vector<int>& nums) {

	}
//};



int main(int argc, char* argv[])
{



	string start = "hit";
	string end = "cog";


	//cout << "Input the string" << endl;
	//string str;

	//cin >> str;

	string str = "aab";





	vector<vector<string>> result;



	Solution s;
	result = s.partition(str);
	//stackTree.push(p->left);
	//stackTree.push(p->right);
	//if (s.isPalindrome(str1))
	//	cout << " True" << endl;
	//else
	//	cout << "false" << endl;
	system("pause");
	return 0;
}
//std::unordered_set<std::string> myset =
//{ "hot", "dot", "dog", "lot", "log" };

//std::cout << "myset contains:";
// for (auto it = myset.begin(); it != myset.end(); ++it)
//std::cout << " " << *it;
//;; std::cout << std::endl;

//TreeNode *root = new TreeNode(1);
//TreeNode *left = new TreeNode(2);
//TreeNode *right = new TreeNode(3);

//root->left = left;
//root->right = right;